Difference between revisions of "2024 AMC 10B Problems/Problem 7"

Revision as of 03:26, 14 November 2024

What is the remainder when $7^{2024}+7^{2025}+7^{2026}$ is divided by $19$ ?

$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 7 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 18$

We can factor the expression as

$7^{2024} (1 + 7 + 7^2) = 7^{2024} (57).$

Note that $57$ is divisible by $19$ , this expression is actually divisible by 19. The answer is $\boxed{\textbf{(A) } 0}$ .

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

@@ Line 10: / Line 10: @@
 <cmath>7^{2024} (1 + 7 + 7^2) = 7^{2024} (57).</cmath>
-Note that <math>57</math> is divisible by <math>19</math>, this expression is actually divisible by 19. The answer is <math>\textbf{(A) } 0</math>.
+Note that <math>57</math> is divisible by <math>19</math>, this expression is actually divisible by 19. The answer is <math>\boxed{\textbf{(A) } 0}</math>.
 ==See also==
 {{AMC10 box|year=2024|ab=B|num-b=6|num-a=8}}
 {{MAA Notice}}